The time period of a soap bubble is T which depneds on pressure p density d and surface tension s then the relation for T is

Options

A. `T alpha p ds`
B. `T alpha p^(-3//2) d^(1//2)s`
C. `T alpha p d^(-2)s^(3//2)`
D. `T alpha p d^2 s^(1//3)`

Answer

To find the relationship for the time period \( T \) of a soap bubble in terms of pressure \( p \), density \( d \), and surface tension \( s \), we can use dimensional analysis. Here’s a step-by-step solution:

### Step 1: Set up the relationship
We assume that the time period \( T \) can be expressed as:
\[ T = k \cdot p^{\alpha} \cdot d^{\beta} \cdot s^{\gamma} \]
where \( k \) is a proportionality constant, and \( \alpha \), \( \beta \), and \( \gamma \) are the powers to be determined.

### Step 2: Write the dimensions of each quantity
- The dimension of time \( T \) is:
\[ [T] = T \]
- The dimension of pressure \( p \) (force per unit area) is:
\[ [p] = \frac{F}{A} = \frac{M L T^{-2}}{L^2} = M L^{-1} T^{-2} \]
- The dimension of density \( d \) (mass per unit volume) is:
\[ [d] = \frac{M}{L^3} = M L^{-3} \]
- The dimension of surface tension \( s \) (force per unit length) is:
\[ [s] = \frac{F}{L} = \frac{M L T^{-2}}{L} = M T^{-2} \]

### Step 3: Substitute the dimensions into the equation
Substituting the dimensions into the equation gives:
\[ [T] = [p]^{\alpha} \cdot [d]^{\beta} \cdot [s]^{\gamma} \]
This translates to:
\[ T = (M L^{-1} T^{-2})^{\alpha} \cdot (M L^{-3})^{\beta} \cdot (M T^{-2})^{\gamma} \]

### Step 4: Expand and group the dimensions
Expanding this, we get:
\[ T = M^{\alpha + \beta + \gamma} \cdot L^{-\alpha - 3\beta} \cdot T^{-2\alpha - 2\gamma} \]

### Step 5: Set up equations based on dimensions
For the dimensions to be consistent, we equate the powers of \( M \), \( L \), and \( T \):
1. For mass \( M \):
\[ \alpha + \beta + \gamma = 0 \quad \text{(1)} \]
2. For length \( L \):
\[ -\alpha - 3\beta = 0 \quad \text{(2)} \]
3. For time \( T \):
\[ -2\alpha - 2\gamma = 1 \quad \text{(3)} \]

### Step 6: Solve the equations
From equation (2):
\[ \alpha = -3\beta \quad \text{(4)} \]
Substituting (4) into (1):
\[ -3\beta + \beta + \gamma = 0 \implies -2\beta + \gamma = 0 \implies \gamma = 2\beta \quad \text{(5)} \]
Substituting (4) and (5) into (3):
\[ -2(-3\beta) - 2(2\beta) = 1 \implies 6\beta - 4\beta = 1 \implies 2\beta = 1 \implies \beta = \frac{1}{2} \]
Using (5):
\[ \gamma = 2 \cdot \frac{1}{2} = 1 \]
Using (4):
\[ \alpha = -3 \cdot \frac{1}{2} = -\frac{3}{2} \]

### Step 7: Write the final relationship
Now substituting \( \alpha \), \( \beta \), and \( \gamma \) back into the original relationship:
\[ T \propto p^{-\frac{3}{2}} \cdot d^{\frac{1}{2}} \cdot s^{1} \]

Thus, the final relation for the time period \( T \) is:
\[ T \propto \frac{s \sqrt{d}}{p^{3/2}} \]